`1.44` gram if titanium `(Ti)` reacted with excess of `O_(2)` and produce `x` gram of non`-` stoichiometric compound `Ti_(1.44)O`. The value of `x` is `:`

Options

A. 2
B. `1.77`
C. `1.44`
D. None of these

Answer

To find the mass \( x \) of the non-stoichiometric compound \( Ti_{1.44}O \) produced from the reaction of 1.44 grams of titanium \( (Ti) \) with excess oxygen \( (O_2) \), we can follow these steps:

### Step 1: Calculate the number of moles of titanium
First, we need to find the number of moles of titanium. The molar mass of titanium \( (Ti) \) is approximately \( 48 \, \text{g/mol} \).

\[ \text{Number of moles of } Ti = \frac{\text{mass of } Ti}{\text{molar mass of } Ti} = \frac{1.44 \, \text{g}}{48 \, \text{g/mol}} = 0.03 \, \text{mol} \]

### Step 2: Determine the number of moles of the compound \( Ti_{1.44}O \)
Since all the titanium reacts to form the compound \( Ti_{1.44}O \), the number of moles of \( Ti_{1.44}O \) will also be \( 0.03 \, \text{mol} \).

### Step 3: Calculate the molar mass of the compound \( Ti_{1.44}O \)
The molar mass of the compound \( Ti_{1.44}O \) can be calculated as follows:

\[ \text{Molar mass of } Ti_{1.44}O = (1.44 \times 48 \, \text{g/mol}) + (1 \times 16 \, \text{g/mol}) = 69.12 \, \text{g/mol} + 16 \, \text{g/mol} = 85.12 \, \text{g/mol} \]

### Step 4: Calculate the mass \( x \) of the compound \( Ti_{1.44}O \)
Now, we can find the mass \( x \) of the compound using the number of moles calculated earlier:

\[ x = \text{Number of moles of } Ti_{1.44}O \times \text{Molar mass of } Ti_{1.44}O \]

\[ x = 0.03 \, \text{mol} \times 85.12 \, \text{g/mol} = 2.5536 \, \text{g} \]

### Step 5: Final calculation
Since we need to express the mass \( x \) in grams, we can round it to two decimal places:

\[ x \approx 2.55 \, \text{g} \]

### Conclusion
The value of \( x \) is approximately \( 2.55 \, \text{g} \).

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